Question Number 191303 by Mingma last updated on 22/Apr/23
Answered by mahdipoor last updated on 22/Apr/23
![∫f(x)=∫xf^′ (x)−(√(2x−x^2 ))=[xf(x)−∫f(x)]−∫(√(2x−x^2 )) ⇒∫_0 ^( 1) f(x)=(1/2)[xf(x)]_0 ^1 −∫_0 ^( 1) (√(2x−x^2 ))= 0−∫_0 ^( 1) (√(2x−x^2 ))](https://www.tinkutara.com/question/Q191306.png)
$$\int{f}\left({x}\right)=\int{xf}^{'} \left({x}\right)−\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }=\left[{xf}\left({x}\right)−\int{f}\left({x}\right)\right]−\int\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[{xf}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }= \\ $$$$\mathrm{0}−\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by cortano12 last updated on 22/Apr/23
![⇒x ((df(x))/dx) = f(x)+(√(2x−x^2 )) ⇒x df(x)=f(x)dx+(√(−(x^2 −2x+1)+1)) dx ⇒∫_0 ^1 x df(x)=∫_0 ^1 f(x)dx+∫_0 ^1 (√(1−(x−1)^2 )) dx ⇒[x f(x)]_0 ^1 −∫_0 ^1 f(x)dx=∫_0 ^1 f(x)dx+∫_0 ^1 (√(1−(x−1)^2 )) dx ⇒−2∫_0 ^1 f(x)dx=(π/4) ⇒∴ ∫_0 ^1 f(x)dx =−(π/8)](https://www.tinkutara.com/question/Q191308.png)
$$\:\Rightarrow\mathrm{x}\:\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}\:=\:\mathrm{f}\left(\mathrm{x}\right)+\sqrt{\mathrm{2x}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{x}\:\mathrm{df}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\sqrt{−\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}\right)+\mathrm{1}}\:\mathrm{dx} \\ $$$$\Rightarrow\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{x}\:\mathrm{df}\left(\mathrm{x}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\sqrt{\mathrm{1}−\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\Rightarrow\left[\mathrm{x}\:\mathrm{f}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\Rightarrow−\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\therefore\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\:=−\frac{\pi}{\mathrm{8}} \\ $$